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Physics-

General

Easy

Question

A wheel of radius 1m rolls forward half a revolution on a horizontal ground. The magnitude to the displacement of the point of the wheel initially in contact with the ground is

$2 π$
$\sqrt{2 π}$
$\sqrt{π^{2} + 4}$
$π$

The correct answer is: $square root of pi to the power of 2 end exponent plus 4 end root$

When wheel rolls half a revolution, the point (P ) of the wheel which is in contact with the ground initially, moves at the top of the wheel as shown.

Horizontal displacement of point P,
$y equals 2 r$
$N e t blank d i s p l a c e m e n t equals square root of x to the power of 2 end exponent plus y to the power of 2 end exponent end root$
$equals square root of open parentheses pi r close parentheses to the power of 2 end exponent plus open parentheses 2 r close parentheses to the power of 2 end exponent end root$
$equals r square root of pi to the power of 2 end exponent plus 4 end root$
$equals square root of pi to the power of 2 end exponent plus 4 end root$

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